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Given a probability space $(\mathbb{R}^k, \mathcal{B}^k, \mathbb{P}_1)$ and a Markov kernel $\mathbb{P}_{1,2}:\mathbb{R}^k\times\mathcal{B}^l\rightarrow\mathbb{R}$ on a measure space $(\mathbb{R}^l, \mathcal{B}^l)$. Then there exists a probability measure $\mathbb{P}$ given by $$\mathbb{P}(A)=\int_{\mathbb{R}^k}\left(\int_{\mathbb{R}^l}\mathbb{1}_{A}(\omega_1, \omega_2)\mathbb{P}_{1,2}(\omega_1, d\omega_2)\right)\mathbb{P}_1(d\omega_2),$$ for $A\in \mathcal{B}^k \otimes \mathcal{B}^l=\mathcal{B}^{k+l}$.

If we now introduce random variables $X_1$ and $X_2$ on $(\mathbb{R}^k, \mathcal{B}^k, \mathbb{P}_1)$ and $(\mathbb{R}^l, \mathcal{B}^l)$, respectively, we can form the random variable $(X_1, X_2)$ defined on $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l}, \mathbb{P})$.

1.) So now I want to write down the distribution of the random variable $X_1$, i.e., $\mathbb{P}_{X_1}$. This is a push forward measure. But my question is, a pushforward measure of which measure? I.e., of $\mathbb{P}_1$ or of $\mathbb{P}$?

2.) Can it maybe be discribed in terms of either one? If so how would it look like if it is defined as the push forward measure of $\mathbb{P}$, which takes as an input sets from $\mathcal{B}^{k+l}$, but $X_1$ is only defined on $\mathbb{R}^k$?

3.) More generally, given a random variable $X$ and a random variable $Y$ on two different spaces $(\mathbb{R}^k, \mathcal{B}^k)$ and $(\mathbb{R}^l, \mathcal{B}^l)$, that have a joint distribution $\mathbb{P}_{X, Y}$ on $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l})$. Then of what measure is $\mathbb{P}_X$ a pushforward measure of? And of what measure is the joint distribution a forward measure of? I.e., do we require a measure $\mathbb{P}$ on the product space $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l})$ and are the marginals and the joint distribution push forward measure of that $\mathbb{P}$? If, so, how does it work for $\mathbb{P}_X$ as the dimensions dont match?

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  • $\begingroup$ I think there is some confusion about some basics here. $X_1$ and $X_2$ map what space to what space exactly? Equivalently, if they take values in some space, which space do they take as input? An answer will also depend on this information, since otherwise the concept of pushforward measure seems lost to me. $\endgroup$ Commented Feb 24, 2023 at 19:40
  • $\begingroup$ Hi and thank you for your comment! So I think that here it does not matter to which space the random variables are mapping, right? Since we are interested in the pushforward measure whcih is defined via the preimage we only need to know the domain and the corresponding sigma algebra. For $X_1$ this is $(\mathbb{R}^k, \mathcal{B}^k, \mathbb{P}_1)$ and for $X_2$ this is $(\mathbb{R}^l, \mathcal{B}^l)$ $\endgroup$ Commented Feb 25, 2023 at 8:56
  • $\begingroup$ No. You start with an arbitrary probability space $(\Omega,\mathscr{F},P)$ and define a rv $X:\Omega\to \mathbb{R}^n$ s.t. $\mu=P\circ X^{-1}$ is the pushforward measure on $(\mathbb{R}^n,\mathscr{B})$. You need two spaces and a measurable map between them to talk about pushforward measure. In your case you have a probability space, measurable maps, but no information on the target space so pushforward makes no sense here. $\endgroup$ Commented Feb 25, 2023 at 10:35

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As in the comments, it seems like there's some confusion here over how to set up your notation. In typical usage, you are interested in one specific probability space $(\Omega, \mathcal F, \mathbb P)$, and all random variables are measurable functions of that space. You could for example set $(\Omega, \mathcal F) = (\mathbb R^{k+l}, \mathcal B^{k+l})$, and define $\mathbb P$ as in your display, though this is by no means required.

You would then define random variables $X_1$ and $X_2$ as measurable functions on $(\mathbb R^{k+l}, \mathcal B^{k+l})$; if $X_1$ only depends on the first $k$ coordinates, and $X_2$ on the last $l$, then those are just some properties the functions happen to satisfy. You wouldn't normally start with some other random variables $X_1$ and $X_2$, defined on two other spaces, and try to copy them over.

To answer your questions, the push-forward measure $\mathbb P_{X_1}$ is then straightforwardly $\mathbb PX_1^{-1}$. If random variables $X$ and $Y$ are defined on two spaces $S$ and $T$, then it does not make sense to say they have a joint distribution on some third space $U$; they're defined on $S$ and $T$.

You could, if you like, define some new random variables $X'$ and $Y'$ on $U$, which have the same marginal distributions as $X$ and $Y$, but these would be different variables, and their push-forward measures would be defined in the usual way.

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  • $\begingroup$ thank you for your answer. I have still some questions: So if I have two random variables $X$ and $Y$ and they are defined on two different spaces $(\mathbb{R}^k, \mathcal{B}^k)$ and $(\mathbb{R}^l, \mathcal{B}^l)$ can I always pretend as if they were defined on a common space $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l})$? If so, does this only hold for spaces over the reals with borel sigma-algebra? And does a joint distribution then always exist? $\endgroup$ Commented Feb 27, 2023 at 8:06
  • $\begingroup$ Maybe the opposite question is also interesting: If I have a joint distribution of some random variables $X$ and $Y$ on a common space $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l})$, does it mean that $X$ and $Y$ will always also be defined on that common space $(\mathbb{R}^{k+l}, \mathcal{B}^{k+l})$? But what happens if $X$ is defined as $X=id_{\mathbb{R}^k}$ and $Y=id_{\mathbb{R}^l}$, i.e., they are the $k$ and $l$ dimensional identity map, respectively? Wouldnt that imply that they are actually defined on $(\mathbb{R}^{k}, \mathcal{B}^{k})$ and $(\mathbb{R}^{l}, \mathcal{B}^{l})$, respectivaely? $\endgroup$ Commented Feb 27, 2023 at 8:12
  • $\begingroup$ Given $X$ defined on $(U, \mathcal U, \mu)$, and $Y$ defined on $(V, \mathcal V, \nu)$, you can always define $X'$ and $Y'$ on $(U \times V, \mathcal U \otimes \mathcal V, \mu \otimes \nu)$ by projection. If $X':U\times V \to U$ is a projection, then it is still defined on $U \times V$; you've misunderstood what "defined on" means. @guest1 $\endgroup$ Commented Feb 27, 2023 at 8:16
  • $\begingroup$ You can always go from a variable $X'$ on $U \times V$ to a variable $X$ on $U$ by integrating out $V$; if the original variable $X'$ was a projection to $U$, then this will have the same effect as ignoring $V$. But $X$ is not $X'$. One is defined on $U \times V$, the other is defined on $U$. @guest1 $\endgroup$ Commented Feb 27, 2023 at 8:28
  • $\begingroup$ thank you! So by projection you mean that the mapping of the random variable still remains the same but we change the domain (where we can however, ignore the inputs that are not needed for the definition of our mapping)? And a question to the new space that you have defined in your first comment. Is $\mu \otimes \nu$ the product measure? Can we always define it when we create a new space like that? And does that mean that any joint distribution of $X'$ and $Y'$ will be independent? $\endgroup$ Commented Feb 27, 2023 at 8:47

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